# Why (Z,+) is cyclic group?

• A
• LagrangeEuler
In summary, the conversation discusses the concept of a cyclic group, specifically in the context of the group of integers under addition. The cyclic subgroup generated by an element is defined as the set of all powers of that element, including negative powers in the case of an additive group. Therefore, both 1 and -1 are generators of the group of integers. It is also mentioned that in an infinite cyclic group, an element and its inverse have the same order and are the only possible generators.

#### LagrangeEuler

I do not understand why ##(Z.+)## is the cyclic group? What is a generator of ##(Z,+)##?
If I take ##<1>## I will get all positive integers. If I take ##<-1>## I will get all negative integers. I should have one element which generates the whole group. What element is this?

LagrangeEuler said:
I do not understand why ##(Z.+)## is the cyclic group? What is a generator of ##(Z,+)##?
If I take ##<1>## I will get all positive integers. If I take ##<-1>## I will get all negative integers. I should have one element which generates the whole group. What element is this?
The positive integers do not form a subgroup, so they cannot be the subgroup generated by the element ##1##. Something is wrong with the definitions you are using.

If ##(G, \cdot)## is a group and ##g\in G##, the cyclic subgroup generated by ##g## is ##\langle g\rangle = \{g^n : n\in \mathbb{Z}\}##, where ##g^0 = e##, ##g^n= g\cdot g\cdots g## (##n## times) if ##n## is a positive integer and ##g^n = g^{-1}\cdot g^{-1}\cdots g^{-1}## (##-n## times) is ##n## is a negative integer.

If ##G## is additive, ##ng## is written in place of ##g^n##. So when ##G## is additive ##\langle g\rangle = \{ng : n\in \mathbb{Z}\}##. In the case ##G = \mathbb{Z}##, the cyclic subgroup ##\langle 1\rangle = \{n\cdot 1 : n\in \mathbb{Z}\} = \{n : n\in \mathbb{Z}\} = \mathbb{Z}##. Similarly ##\langle -1\rangle = \mathbb{Z}##. So indeed, ##\mathbb{Z}## is cyclic, and both ##-1## and ##1## are generators of the group.

• topsquark and SammyS
Euge said:
If ##G## is abelian,

• topsquark and SammyS
PeroK said:
Corrected, thanks!

Can I add, as a non-mathematician….

If a finite group is cyclic, the group can be generated by a single element.

But if an infinite group (such as integers under addition) is cyclic, the group can be generated by a single element or its inverse. So in the present case we can use 1 or -1.

• nuuskur, topsquark and SammyS
Steve4Physics said:
Can I add, as a non-mathematician….

If a finite group is cyclic, the group can be generated by a single element.

But if an infinite group (such as integers under addition) is cyclic, the group can be generated by a single element or its inverse. So in the present case we can use 1 or -1.
A group element has the same order as its inverse, so for any group ##G## and ##g\in G##, the cyclic subgroup ##\langle g\rangle = \langle g^{-1}\rangle##.

• Greg Bernhardt, topsquark and PeroK
A fun exercise is proving in an infinite cyclic group that ##g## and ##-g## are the only generators, which is not true for finite groups - e.g. the cyclic group of order ##p## for ##p## prime has every element as a generator except for the identity.

• topsquark and Euge